One newly discovered light particle has a mass of m and property q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a property Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? Am,m2
One newly discovered light particle has a mass of m and property q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a property Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? Am,m2
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One newly discovered light particle has a mass of m and property q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a property Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the
heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle?
Am¡m2
Consider a new expression for gravitation potential energy as: PEgrav=
, where A is a constant, m, and m, are the masses of the two objects, and r is the distance between them.
Moreover, the new particle has an additional interaction with the heavy particle through the following force expression
1
Fnew =
4TE0 r2
qQ
where Eg is a constant that is read as epsilon subscript 0, q and Q are their new properties, r is the distance between the new particle and the heavy particle.
Solution:
We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem.
To avoid the complexity of vector solution, we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle.
Let us first name the lighter particle as object 1 and the heavy particle as object 2.
Through work-energy theorem, we will take into account all of the energy of the two-charged particle system before and after traveling a certain distance as
KE1F + KE2F + PEgravf + Velasticf + Unewf = KE11 + KE2¡ + PEgravi +
+ Unewi
Since the heavy particle remains fixed, before
after the motion of the lighter particle, it does not have any velocity, moreover, there is no spring involved, so
KE1F +
+ Unewf =
+ Unewi
(Equation 1)
For all energies, we know the following
1
KE =
mv](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3333b9da-002b-475c-ab16-c7c2fcc70c27%2F84946d71-814d-4efc-936b-b1ff2e6753db%2F5xjri1o_processed.png&w=3840&q=75)
Transcribed Image Text:Problem
One newly discovered light particle has a mass of m and property q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a property Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the
heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle?
Am¡m2
Consider a new expression for gravitation potential energy as: PEgrav=
, where A is a constant, m, and m, are the masses of the two objects, and r is the distance between them.
Moreover, the new particle has an additional interaction with the heavy particle through the following force expression
1
Fnew =
4TE0 r2
qQ
where Eg is a constant that is read as epsilon subscript 0, q and Q are their new properties, r is the distance between the new particle and the heavy particle.
Solution:
We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem.
To avoid the complexity of vector solution, we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle.
Let us first name the lighter particle as object 1 and the heavy particle as object 2.
Through work-energy theorem, we will take into account all of the energy of the two-charged particle system before and after traveling a certain distance as
KE1F + KE2F + PEgravf + Velasticf + Unewf = KE11 + KE2¡ + PEgravi +
+ Unewi
Since the heavy particle remains fixed, before
after the motion of the lighter particle, it does not have any velocity, moreover, there is no spring involved, so
KE1F +
+ Unewf =
+ Unewi
(Equation 1)
For all energies, we know the following
1
KE =
mv
![Am¡m2
PEgrav =
r
U elastic = kx
kx²
Unew = (1/
where in we have
m1 = m, m2 = M, q1 = q and q2 = Q
By substituting all these to Equation 1 and then simplifying results to
= sgrt(
2 + ( (
m ) -
) - (1/x
) ) +
V
Take note that capital letters have different meaning than small letter variables/constants.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3333b9da-002b-475c-ab16-c7c2fcc70c27%2F84946d71-814d-4efc-936b-b1ff2e6753db%2Fyb9qdxh_processed.png&w=3840&q=75)
Transcribed Image Text:Am¡m2
PEgrav =
r
U elastic = kx
kx²
Unew = (1/
where in we have
m1 = m, m2 = M, q1 = q and q2 = Q
By substituting all these to Equation 1 and then simplifying results to
= sgrt(
2 + ( (
m ) -
) - (1/x
) ) +
V
Take note that capital letters have different meaning than small letter variables/constants.
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